[Stoves] A few stove questions

[Robert Penn Taylor]

Hi Frank,

Responses interleaved. Sorry for the length; I kept it as short as I 
could while still doing justice to the question.

frank wrote:
> Dear stovers,
> I have a few questions:
> 
> Does water at boiling have 1561 J/g energy? That is 4.186 J/C/g starting 
> at -273 deg C (0 deg K).and going to + 100 deg C?
> 

Unfortunately it's not so simple as that. For starters, it depends on 
what you mean by "energy in the water". Without going into detail about 
it, I'll just point out that you may want to know about internal energy, 
and you may want to know about enthalpy. Any thermodynamics textbook you 
can dig up should adequately explain the difference between the two. In 
dealing with stoves and energy calculations, enthalpy is usually the 
property we want.

Let's assume you're heating up water at constant pressure, which is the 
appropriate assumption to make if you're not using a pressure cooker. If 
you look at the difference in enthalpy between the final water 
temperature and the initial water temperature, you'll know how much 
energy per unit mass is required to raise the water through that 
temperature. The easiest way to do this is to look up the enthalpies in 
a table, such as the ones that can be generated at 
http://webbook.nist.gov/chemistry/fluid/.

If you're really interested in being able to calculate it in a way 
similar to what you were trying by doing 4.186 * 373, email me and I can 
explain where and why that process went wrong. It would make this post 
way too long.

> If you need an additional 2260 J/g after the water reaches 100 deg C to 
> bring it to a boil that is like bringing the temperature to (2260 / 
> 4.186) 540 deg. C but water does not go above 100 deg C (?). How is this 
> energy stored until it builds up to boiling point? Where is it?
> 

It's stored as internal "vibrations" in the molecule. If you think of 
the two H atoms as being connected to the O atom via springs, imagine 
that once you get to this temperature you start compressing the springs, 
adding energy without increasing the temperature. To be a little more 
specific, at 1 atm, 2087 J/g goes into internal energy (the springs) and 
an additional 169 J/g goes into creating the volumetric expansion that 
occurs when you go from liquid to gas.

> If measuring the temperature along a very tall stack is there a place 
> where the temperature drops to 100 deg C (but still has the +2260 J/g 
> energy) and the water condenses releasing the +2260 J/g so the 
> temperature goes up?
> 

I don't think so. There will just be a long section of stack where the 
temperature stays at 100 C.

> If the fire under Lannys stove is of the size that before the heat 
> exits  it is reduced (fins and length) to below 100 deg C.  will the 
> released heat be included in heating the water and we should go by the 
> HHV and not the LHV of the wood? And is this a goal of producing the 
> 'perfect' stove?
> 

The released heat will certainly help heat the water -- as long as the 
water in the pot is below boiling too. In my opinion, we should always 
be using the HHV of the fuel when stating first-law efficiencies, and 
just come to grips with the fact that 100% efficiency is unachievable in 
a heat transfer device. Using the LHV hides this issue and muddies the 
concepts involved.

-Penn

-- 

Robert Penn Taylor
Graduate Research Assistant
Department of Mechanical Engineering
Iowa State University
(515) 294-5311